Codeforces Round #511 (Div. 2) C. Enlarge GCD (思维题)

原题地址:http://codeforces.com/contest/1047/problem/C

题意:给出n个数字,现在询问你最少需要删除多少个数字使得剩余数字的gcd会增大,如果不能不能就输出-1

思路:我们考虑先求出所有数字的gcd,然后再让每一个数字除这个gcd,那么除完之后的序列的gcd必定为1.那么如果这些数字全部是1,那么必然不能使得gcd变大.否则的话,我们只需要记录每一个数字的有哪些质因数,然后去寻找出现最多的那个质因子就行了.

#include <bits/stdc++.h>
#define eps 1e-8
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define CLR(x,y) memset((x),y,sizeof(x))
#define fuck(x) cerr << #x << "=" << x << endl

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int seed = 131;
const int maxn = 3e5 + 5;
const int mod = 1e9 + 7;
int n;
int a[maxn];
const int mx = 15e6 + 5;
int noprime[mx], pcnt, p[mx]; //p存的是质数
void getprime(int N) {
    pcnt = 0;
    memset(noprime, 0, sizeof(noprime));//1表示是质数
    noprime[0] = noprime[1] = 1;
    for (int i = 2; i < N; ++i) {
        if (!noprime[i]) {
            noprime[i] = i; //表示数字i是第几个质数,3会是1,需要特判
            p[pcnt++] = i;//p是存的所有质数
        }
        for (int j = 0; j < pcnt && i * p[j] < N; ++j) {
            noprime[i * p[j]] = p[j];//i*p[j]有一个质因子是p[j]
            if (i % p[j] == 0)break;
        }
    }
}
int num[mx];
int main() {
    scanf("%d", &n);
    getprime(15e6 + 3);

    int flag = 0;
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);

    }
    int gcd = 0;
    for (int i = 1; i <= n; i++) gcd = __gcd(gcd, a[i]);
    for (int i = 1; i <= n; i++) {
        a[i] /= gcd;
        if (a[i] != 1) flag = 1;
    }
    for (int i = 1; i <= n; i++) {
        while (a[i] != 1) {
            int x = noprime[a[i]];
            num[x]++;
            while (a[i] % x == 0) {
                a[i] /= x;
            }
        }
    }
    if (!flag) {
        printf("-1\n");
        return 0;
    }
    int MIN = INF;
    for (int i = 0; i < mx; i++) {
        MIN = min(MIN, n - num[p[i]]);
    }
    printf("%d\n", MIN);
    return 0;
}